Complex analysis of $f(z) = 2 y^2 \sin x − i y^3 \cos x$

Question

I want to find the domain of definition, the domain of derivability and the domain of analyticity of the function $f(z) = 2 y^2 \sin x − i y^3 \cos x$

The domain of definition is $\mathbb C$ but I don't know how to find the otres two in this function.

Answer 1

$$u=2y^2\sin x,v=-y^3\cos x$$ $$\frac{\partial u}{\partial x}=2y^2\cos x,\frac{\partial v}{\partial y}=-3y^2\cos x$$ $$\frac{\partial u}{\partial y}=4y\sin x,-\frac{\partial v}{\partial x}=-y^3\sin x$$ Suppose $\sin x=0$, then $\cos x$ can be cancelled in the first relation, leaving $y=0$ as the only solution. Otherwise, $\sin x$ can be cancelled in the second relation, giving $4y=-y^3$; its only solution is $y=0$.

Hence $f$ is differentiable only on the real axis and not analytic anywhere.

Answer 2

Let's check when the Cauchy-Riemann equations hold: $$u=2y^2\sin x, v=-y^3\cos x$$ $$u_x=2y^2\cos x, v_y=-3y^2\cos x$$ $$u_y=4y\sin x, -v_x=-y^3\sin x$$ A necessary condition for being differentiable is for the CR equations to hold - $u_x=v_y,u_y=-v_x$. From this, you get a strong restriction on the possible points where $f$ is differentiable (certainly not everywhere). Moreover, it's easy to see that $f$ cannot be differentiable on any open set, so it's nowhere analytic.